Download Midterm, Version 1, Solutions

SOLUTIONS TO MIDTERM
VERSION 1
1) The random variable X= # Likely Voters for the Democrat (in the sample of
size 3) has a binomial distribution with n=3, p=.4. The Democrat is preferred
if either 2 or 3 in the sample support the Democrat. So we need
 3
 3
p(2)  p(3)    p 2 q    p 3  3(.4 2 )(1  .4)  .4 3  .352.
 2
 3
Answer is C.
2) We first need to determine n and p. Since the mean is np = 2 and the variance
is npq = 1, we can divide the second equation by the first to conclude that q =
.5, and therefore p = .5 and then the first equation implies that n=4. If X is
binomial with n=4, p=.5, we need Prob{X ≥ 1} = 1 −Prob{X=0} =
1  .5 4  .9375. Answer is D.
3) If X=Profit, then E[X] = (50000)(.6) + 0(.4) = 30,000. Thus,
Var(X)=(50000−30000)2(.6)+(0−30000)2(.4) = 600,000,000 so the standard
deviation is the square root of the variance, $24,494.90. Answer is A.
4) They cannot be independent since P(A|B) = P(A  B) / P(B) = 0 / P(B) = 0
which is not equal to P(A). Answer is A.
5) Note that σ = 2 and n = 50, so we get  x =  / n  2 / 50  .2828.
Answer is B.
6) Since the variance of the binomial distribution is npq the standard deviation is
  5(.4)(.6)  1.095. Answer is D.
7) If X is the normal random variable, then P(X < 0) = P{Z < (0−2)/4} = P{Z <
−.5} = .5 − .1915 = .3085. Answer is A.
8) If Xi is the profit for the ith round, we want = Prob (X1 + …+X50 ≥10) =
Prob ( X ≥ .2). By the Central Limit Theorem, we can treat X as having a
normal distribution with mean  x    .014 and standard deviation

1.00
= .1414. Thus, we get
x 

n
50
Prob { X ≥ .2}=Prob{Z≥(.2+.014)/.1414}=Prob{Z≥1.51}=.5−.4345=.0655.
Answer is D.
9) From Table 5, Prob{Z < −1.5}=.5−.4332 = .0668. Answer is A.
10) If A={Big Change Today} and B={Big Change Tomorrow}, then
P(A)=P(B)=.04. From the statement of the problem, we have P(B|A)=.1. Note
that A, B are not independent. Using the rule of complements and the general
multiplication rule, we conclude that the probability of no big change today
and no big change tomorrow is
P( A  B )  1  P( A  B)  1  [ P( A)  P( B)  P( A) P( B | A)]
 1  [.04  .04  .04(.1)]  .924. Answer is D.
11) If X is the average number of fans for the next n=30 days, and Xi is the
number of fans for the ith then Xi has mean μ=1.5 and standard deviation σ
=1. Thus, X has mean  x    1.5 and standard deviation

1
x 

 .1826 . Therefore, by the Central Limit Thorem,
n
30
Prob{Win 50 free plays} = Prob{Total number of fans ≥ 40} =
Prob{ X ≥1.333} = Prob{Z ≥(1.333−1.5)/.1826} =
Prob{Z≥−.91}=.5+.3186=.8186. Now, let Y=Number of Free plays. We
have shown that Y will equal 50 with probability .8186, and Y will equal
zero with probability 1−.8186 = .1814. Thus, E[Y]=50(.8186)+0 = 40.93.
Answer is B.
12) X= #Profitable Years of next 10 has a binomial distribution with n=10, p=.75.
We need Prob{X≥9} =
10 
10 
p(9)  p(10)   (.759 )(.251 )    (.7510 )  (10)(.759 )(.251 )  .7510  .2440.
9
10 
Answer is D.
13) Since the z-score is standard normal, we have Prob(Z>1)=.5−.3413 = .1587.
Answer is A.
14) X=Total Number of Heads in 150 tosses is binomial with n=150, p=.5, so the
standard deviation of X is npq  (150)(.5)(.5)  6.124. Answer is C.
15) The general addition rule is P(AB)=P(A)+P(B)−P(A  B). Comparing this
with the formula given in the problem, we conclude that P(A B)=P(A)P(B),
which can only happen if A and B are independent. Answer is B.